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I'm so confused this question is very different from the other hydrostatic force questions and I think I am misunderstanding the question. enter image description here

I am primarily concerned with 15 because I somehow managed to get 13 correct. Am I correct in assuming this shape has a z-axis depth that is not given and does not matter? I am used to modeling Work = $\int_a^b Force$ (Force is in most cases weight density * height * area of cross-section)

I could really use some pointers on how to integrate this. I realize the sides can be modeled with the equations y=$\frac 83$ x

My attempt: $\int_0^5 62.4*2* {\frac 83} x*(5-x) dx$ and I got 6933 when the correct answer is ~3705. I don't know where I went wrong..

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  • $\begingroup$ You forgot to enclose the TeX in $'s. $\endgroup$ May 19, 2015 at 23:53

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Decide which direction is "up" and which axis points in which direction. Evidently the narrowest angle of the triangle is "up" and the side of length $6$ is "down".

Based on the formula $y = \frac83 x$, I thought you had decided that the $y$-axis points either up or down (not clear which; this would be important) and the $x$-axis points to the right or left (it does not really matter which). Looking at your integral, however, I began to suspect that you decided your $x$-axis would point straight up, with $x=0$ at the bottom of the tank (so that the depth at any point is $5 - x$) and that the $y$-axis points left or right.

If you meant for the $x$-axis to point upward then you have the wrong formula for the sides of the triangle. Both the slope and the $y$-intercept are incorrect. You did remember to multiply by $2$ in order to count both sides of the triangle (left or right of the centerline), so I think if you replace $\frac83 x$ with the correct formula for $y$ along one side of the triangle then you will be OK.

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  • $\begingroup$ Thank you, I decided to 'slice' sections horizontally and integrate them from bottom to top, so the 5-x is correct. Now that I think about it, if I do this method, the slope is different from either side. it is negative $\frac 83$ x on the right so I couldn't slice it horizontally right? $\endgroup$
    – Obliv
    May 20, 2015 at 11:43
  • $\begingroup$ It seems that when I change the bounds to $\int_3^8$ and model height as x-3 I get the correct answer. I'm going to try and absorb this thank you for your help. $\endgroup$
    – Obliv
    May 20, 2015 at 11:58
  • $\begingroup$ If you make the $x$ axis point down with the zero at the top vertex of the triangle, so $x=8$ at the bottom, then indeed $x-3$ gives the depth of the water ($8-3=5$ at the bottom, which is $5$ feet below the surface) and $\int_3^8$ integrates from the surface to the bottom of the tank--and that is fine. $\endgroup$
    – David K
    May 20, 2015 at 12:34

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